Answer:
0.5015 = 50.15% probability that it came from manufacturer A.
Explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = (P(A \cap B))/(P(A))](https://img.qammunity.org/2022/formulas/mathematics/college/r4cfjc1pmnpwakr53eetfntfu2cgzen9tt.png)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Defective
Event B: From manufacturer A.
Probability a unit is defective:
2% of 43%(from manufacturer A)
1.5% of 57%(from manufacturer B). So
![P(A) = 0.02*0.43 + 0.015*0.57 = 0.01715](https://img.qammunity.org/2022/formulas/mathematics/college/lz9eoi8gzemhhuj7o5qq3pwp240yz95tkv.png)
Probability a unit is defective and from manufacturer A:
2% of 43%. So
![P(A \cap B) = 0.02*0.43 = 0.0086](https://img.qammunity.org/2022/formulas/mathematics/college/uvt5dr2638vz2z6jrl7cluj4rnt73jqfw4.png)
What is the probability that it came from manufacturer A?
![P(B|A) = (P(A \cap B))/(P(A)) = (0.0086)/(0.01715) = 0.5015](https://img.qammunity.org/2022/formulas/mathematics/college/fuiyy921bw4lpg5xh87p44lcbq7c8e1wis.png)
0.5015 = 50.15% probability that it came from manufacturer A.